What I would like to know is how the 'fix' function could be used to find the fixed point of a function like ( \n -> 2*n ).
Olaf and Lennart said a little about least fixpoints, but little about what makes a fixpoint least. Olaf suggested looking up papers on domain theory or denotational semantics to understand it better. But the idea is really simple: just think of it as an ordering on information content. Bottom (_|_) contains NO information, and is thus the least value in any domain. The integers 1, 2, 3, etc. contain the same "amount" of information, but the information in each case is different -- thus these values are incomparable in the information ordering. So even though \n -> n has all these values as fixpoints, the LEAST one is _|_. As for \n -> 2n, even though 0 is a fixpoint, the LEAST one is _|_. And fix will find these for you, in the sense that fix applied to each of these functions leads to non-termination, and non-termination is equivalent to bottom, since it yields NO information. (The fact that in some implementations the stack or heap will blow up is an implementation artifact.)
If it isn't possible, can someone explain the crucial difference between n! and (\n -> 2*n) which allows n factorial to be found via the fix point method and not zero.
fix is able to find the fixpoint of: \f -> \n -> if (n==0) then 1 else n*f(n-1) This fixpoint is a FUNCTION, not the factorial value itself. So they are very different. Which begs the question: can fix find fixpints that are not functions, and not bottom? Consider this well-known example: ones = 1 : ones We can write this as: ones = (\wons -> 1 : wons) ones and thus: ones = fix (\wons -> 1 : wons) Try this in Hugs, and see what happens when you type "take 10 ones" at the prompt. -Paul